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Compund Angle Identities and proof

  1. Feb 20, 2014 #1
    1. The problem statement, all variables and given/known data
    prove using the compound angle identies, proove the following:
    [tex]
    \frac{sin(A-B)}{cos(A)cos(B)}+\frac{sin(B-C)}{cos(B)cos(C)}+\frac{sin(C-A)}{cos(C)cos(A)}=0
    [/tex]

    2. Relevant equations
    n/a


    3. The attempt at a solution
    I resolved it to
    [tex]
    \frac{sin(A)cos(B)-cos(A)sin(B)}{cos(A)cos(B)}+\frac{sin(B)cos(C)-cos(B)sin(C)}{cos(B)cos(C)}+\frac{sin(C)cos(A)-cos(C)sin(A)}{cos(C)cos(A)}=0
    [/tex]

    And using wolfram alpha I found that the first term resolves to tan(A)-tan(B) and then I can see how it equals zero as all the tans would cancel out.

    But I have no idea how each term simplifies to that.

    Any help is really appreciated.

    thanks.
     
  2. jcsd
  3. Feb 20, 2014 #2

    CAF123

    User Avatar
    Gold Member

    After you have expanded the numerators with the addition formulae (which you have done), you can simplify each fraction by taking out a factor.
     
  4. Feb 20, 2014 #3

    Mentallic

    User Avatar
    Homework Helper

    In each fraction, you have something of the form

    [tex]\frac{a+b}{c}[/tex]

    You should always check if splitting up the fraction into

    [tex]\frac{a}{c}+\frac{b}{c}[/tex]

    could help you in any way. In this case, it does.
     
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